Assuming that the equations define $x$ and $y$ implicitly as differentiable functions $x=f(t), y=g(t)$, find the slope of the curve $x=f(t)$, $y=g(t)$ at the given value of $t$.
\[
x=t^{5}+t, y+4 t^{5}=4 x+t^{3}, t=2
\]

Step 1 ：Given the equations \(x=t^{5}+t\) and \(y+4 t^{5}=4 x+t^{3}\) at \(t=2\), we are asked to find the slope of the curve at this point.

Step 2 ：We can find the slope of the curve at a given point by taking the derivative of y with respect to x. However, since we have the functions in terms of t, we can use the chain rule to find dy/dx. The chain rule states that dy/dx = (dy/dt) / (dx/dt).

Step 3 ：First, we need to find dy/dt and dx/dt. We have \(x = t^{5} + t\) and \(y = t^{3} + 4t\).

Step 4 ：Taking the derivative of x with respect to t, we get \(dx/dt = 5t^{4} + 1\).

Step 5 ：Taking the derivative of y with respect to t, we get \(dy/dt = 3t^{2} + 4\).

Step 6 ：Using the chain rule, we find that the slope of the curve is given by \(dy/dx = (dy/dt) / (dx/dt) = (3t^{2} + 4)/(5t^{4} + 1)\).

Step 7 ：Substituting \(t=2\) into the equation for the slope, we find that the slope of the curve at \(t=2\) is \(16/81\).

Step 8 ：Final Answer: The slope of the curve \(x=f(t)\), \(y=g(t)\) at \(t=2\) is \(\boxed{\frac{16}{81}}\).